Question

The slip of a 3-\(\phi\), 50 Hz, induction motor fed by 3-\(\phi\), 50 Hz inverter is 's' at fundamental frequency. At \(5^{\text{th}}\) harmonic frequency, the harmonic slip is nearly equal to

• A. \(5s \)
• B. \(\frac{s}{5} \)
• C. \(1.2 \)
• D. \(1 \)

Hint:

General formula for the harmonic slip of an induction motor: - For forward rotating harmonics (\(n = 7, 13\)): \( s_n = \frac{n - 1 + s}{n} \approx 1 - \frac{1}{n} \) - For backward rotating harmonics (\(n = 5, 11\)): \( s_n = \frac{n + 1 - s}{n} \approx 1 + \frac{1}{n} \) For the $5^{\text{th}}$ harmonic: \( s_5 \approx 1 + \frac{1}{5} = 1.2 \). This short approximation rule protects you from doing long derivation steps during exams!

Answer 1

The Correct Option is C

Concept: Inverter supplies used in industrial drives contain unwanted higher-order voltage and current harmonics. For a standard three-phase system, the harmonic orders present are given by $n = 6k \pm 1$ (where $k = 1, 2, 3 \ldots$). This results in $5^{\text{th}}, 7^{\text{th}}, 11^{\text{th}}, 13^{\text{th}}$ harmonics. Crucially, these harmonics establish magnetic fields inside the air gap that rotate at distinct speeds and directions:
• Harmonic components of order \( n = 6k+1 \) (like 7, 13) rotate in the forward direction (same direction as the fundamental field).
• Harmonic components of order \( n = 6k-1 \) (like 5, 11) rotate in the backward (reverse) direction. The synchronous speed of the $n^{\text{th}}$ harmonic field is $n \cdot N_s$. Because the $5^{\text{th}}$ harmonic rotates in reverse, its synchronous speed relative to the stator is vectorially equal to $-5N_s$.

Step 1: Setting up equations for the fundamental slip.
Let $N_s$ be the fundamental synchronous speed and $N$ be the actual rotor running speed. The fundamental slip $s$ is defined by: \[ s = \frac{N_s - N}{N_s} \quad \Rightarrow \quad N = N_s(1 - s) \quad \cdots (1) \]

Step 2: Formulating the equation for the 5th harmonic slip.
The $5^{\text{th}}$ harmonic field has a synchronous speed magnitude of $5N_s$ moving in the reverse direction. Therefore, the relative slip formula for this $5^{\text{th}}$ harmonic field ($s_5$) is: \[ s_5 = \frac{-5N_s - N}{-5N_s} = \frac{5N_s + N}{5N_s} \]

Step 3: Substituting fundamental parameters into the harmonic slip equation.
Substitute the value of $N$ from Equation (1) into our expression for $s_5$: \[ s_5 = \frac{5N_s + N_s(1 - s)}{5N_s} \] Factoring out $N_s$ from both the numerator and denominator: \[ s_5 = \frac{N_s [5 + (1 - s)]}{5N_s} = \frac{6 - s}{5} \] Splitting the fraction into separate terms: \[ s_5 = \frac{6}{5} - \frac{s}{5} = 1.2 - 0.2s \]

Step 4: Evaluating the approximation for normal running conditions.
Under standard full-load operating conditions, an induction motor runs highly efficiently, meaning the fundamental slip $s$ is extremely small (typically $s \approx 0.02$ to $0.05$). Because $s \ll 1$, the term $0.2s$ becomes small enough to ignore: \[ s_5 \approx 1.2 \] This precisely matches option (3). Hence, the correct choice is option (3).